A star's apparent magnitude is most closely related to which of the following? Select all that apply.a. radius.
b. luminosity.
c. distance from Earth.
d. absolute magnitude. E. surface temperature.

Answers

Answer 1
Answer:

Answer:

Luminosity

Explanation:

The apparent magnitude is a measure of the star's apparent luminosity, the apparent luminosity is directly related to the star's apparent magnitude.

The star's luminosity is the measure of the star's brightness and this diminishes with distance, the luminosity of a star also depends on the size of the star, this means that a star's magnitude and luminosity are directly related.

Answer 2
Answer: If you want the "most", then you can't have more than one.

A star's apparent magnitude is determined by both its intrinsic luminosity
and its distance from us.

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How does the concentration of a hydrogen ion affect the ph levels of a substance?

Answers

Acidic and basic are two extremes that describe chemicals, just like hot and cold are two extremes that describe temperature. Mixing acids and bases can cancel out their extreme effects, much like mixing hot and cold water can even out the water temperature. A substance that is neither acidic nor basic is neutral. 

The character of acidic, basic and neutral is defined by the concentration of hydrogen ions [H+](mol/L). A solution with a concentration of hydrogen ions higher than 10-7mol/L is acidic, and a solution with a lower concentration is alkaline (another way to say basic). Using the formula, pH=-log[H+], a pH of 7 is neutral, a pH less than 7 is acidic, and a pH greater than 7 is basic. As one can see from this formula, ten times a given concentration of hydrogen ions means one unit lower in terms of pH value (higher acidity), and vice versa. 

Pure water is neutral, with a pH of 7.0. When chemicals are mixed with water, the mixture can become either acidic or basic. Vinegar and lemon juice are acidic substances, while laundry detergents and ammonia are basic. 

Chemicals that are very basic or very acidic are called "reactive." These chemicals can cause severe burns. Automobile battery acid is an acidic chemical that is reactive. Automobile batteries contain a stronger form of some of the same acid that is in acid rain. Household drain cleaners often contain lye, a very alkaline chemical that is reactive. 

Charge Q = +4.00 mC is distributed uniformly over the volume of an insulating sphere that has radius R = 5.00 cm. What is the potential difference between the center of the sphere and the surface of the sphere?

Answers

The potential difference between the center of the sphere and the surface of the sphere will be PD=3.6 x 10^8 V

What will be the potential difference?

It is given that:-

Charge Q = +4.00 mC  or  4*10^(-3) c

insulating sphereradius R = 5.00 cm = 0.05 m

The formula for the potentialdifference at the surface of the sphere is given by

V_(s) =(K* Q)/(r) =(9*10^(9) * 4*10^(-3) )/(0.05)=7.2*10^(8) V  

The formula for the potentialdifference at the center of the sphere  

Vc_{} =(3)/(2) * V_(s)

V_(c) = 1.5*7.2*10^(8) v = 10.8*10^(8 ) V      

The difference between the potential difference will be  

V=V_(c) -V_(s)  = 10.8*10^(8) =3.6*10^(8) V

Thus the potential difference between the center of the sphere and the surface of the sphere will be PD=3.6 x 10^8 V

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Answer:

3.6 x 10^8 V

Explanation:

Q = 4 m C = 4 x 10^-3 C

r = 5 cm = 0.05 m

The formula for the potential at the surface is

Vs = K Q / r = (9 x 10^9 x 4 x 10^-3) / 0.05 = 7.2 x 10^8 V

The formula for the potential at the centre is

Vc = 3/2 Vs

Vc = 1.5 x 7.2 x 10^8 V = 10.8 x 10^8 V

The difference in potential is

V = Vc - Vs = 10.8 x 10^8 - 7.2 x 10^8 = 3.6 x 10^8 V

If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?4.0 x 10^-1 A
1.0 x 10^2 A
1.0 x 10^-10 A
4.0 x 10^0 A

Answers

Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
I = 4 A

Q= I×t

Where Q is charge in Coulombs C,

I is current in Amperes A,

t is in seconds s.

The rate of current flow is defined as the charge passing through a point in the circuit per second.

Rearranging formula gives I=(Q)/(t)

⇒I = (2.0*10^-4)/(5.0*10^-5) = 4 Amperes

∴ current flow = 4.0×10^0 A

Where is the field of each charge the strongest ?(A. positive charge arrows pointed out)
(B.negative charge arrows pointed inward)

Answers

First we have to agree on what you mean by "strongest".

To me, "strongest" means the place where it exerts the greatest
force on another charge that happens to be passing by.

In both cases ... 'A' and 'B' ... that's going to be right smack on top
of the charge itself, or as close to it as you can get.

On drawings, it's the spot where the field 'lines' are closest together.

It makes no difference whether the original charges are positive or negative.

a person runs 27.0km west then turns around and runs 13.0km east what's the distance and displacement ?

Answers

Answer:

See below

Explanation:

Distance = 27 + 13 = 40 km

Displacement = 27 - 13 = 14 km

A 0.145-kg baseball pitched at 37.5 m/s is hit on a horizontal line drive straight back toward the pitcher at 57.0 m/s . If the contact time between bat and ball is 2.50×10−3 s , calculate the average force between the ball and bat during contact.

Answers

Answer:

The average force between the ball and bat during contact is 1131N.

Explanation:

The average force can be found employing Newton's second law:

F = ma (1)

Where F is the force, m is the mass and a is the acceleration.

The acceleration can be determined employing the equations for a Uniformly Accelerated Rectilinear Motion:

v_(f) = v_(i) + at  (2)

Where v_(f) is the final velocity, v_(i) is the initial velocity and t is the time.

Then, a will be isolated from equation 2:

a = (v_(f)-v_(i))/(t) (3)

Replacing equation (3) in equation (1) it is gotten:

F = m(v_(f)-v_(i))/(t)

F = (0.145Kg)((57.0m/s-37.5m/s))/((2.50x10^(−3)s))

F = (0.145Kg)((19.5m/s))/((2.50x10^(−3)s))

F = 1131Kg.m/s{2}

But 1N = 1Kg.m/s^(2), therefore:

F = 1131N

So, the average force between the ball and bat during contact is 1131N.